Globally Lipschitz implies solutions exist for all time

I understand when you have a locally lipschitz functiion $f$ for the IVP you are guaranteed the existence and uniqueness of a solution. However this solution may not exist for all time. Why is it that with a globally lipschitz function solutions do exist for all time?

The not so classical proof allows this. Define on the space $C(\Bbb R\to \Bbb R^n)$ of continuous functions the norm $$ \|y\|_L=\sup_{x\in\Bbb R} e^{-2L\,|x|}·\|y(x)\| $$ and consider the space of continuous functions that have a finite norm with this formula. This space contains the constant functions.

Then the Picard operator $$ (P(y))(x)=y_0+\int_0^x f(y(t))\,dt $$ is contractive, since (restrict for the moment to $x>0$, $x<0$ works exactly the same) \begin{align} e^{-2L\,|x|}\left\|(P(y_2))(x)-(P(y_1))(x)\right\| &\le \int_0^x e^{-2L(x-t)}·e^{-2Lt}·L·\|y_2(t)-y_1(t)\|\,dt \\ &\le L·\int_0^x e^{-2L(x-t)}\,dt·\|y_2-y_1\|_L \\ &= \frac12(1-e^{-2L\,x})·\|y_2-y_1\|_L \end{align} so that finally $$ \left\|P(y_2)-P(y_1)\right\|_L\le \frac12·\|y_2-y_1\|_L $$ Now everything you need is supplied by applying the Banach fixed point theorem and the fact that the image of $P$ is actually in $C^1(\Bbb R\to\Bbb R^n)$.

If $f$ is locally Lipschitz and you are looking to solve $u' = f(u)$ with $u(0) = u_0$, the usual procedure allows you to prove existence (and uniqueness) on an interval $[0,t_0]$ where $t_0$ depends on the Lipschitz constant of $f$ but not on $u_0$.

As long as the Lipschitz constant is global you can proceed to finding a solution on $[t_0,2t_0]$, and then $[2t_0,3t_0]$, and so on through all positive times.